# Problem of the Week #43 - March 25th, 2013

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Use contour integration to show that
$\int_{-\infty}^{\infty}\frac{e^{-2\pi i x\xi}}{(1+x^2)^2}\,dx = \frac{\pi}{2}(1+2\pi|\xi|)e^{-2\pi|\xi|}$
for all $\xi$ real.

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Hint:
WLOG, suppose that $\xi\geq 0$ (this way, you don't have to worry about the absolute values for the time being). Then consider using the lower half circle as the contour for this integral.

#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. Here's my solution below.

WLOG, suppose that $\xi\geq 0$ (so we don't have to worry about absolute values for the time being). Let the closed contour $\Gamma$ be the lower half circle. Let $f(z)=\frac{e^{-2\pi i z\xi}}{(1+z^2)^2}$. Clearly, $f(z)$ has two poles of order $2$ at $z=i$ and $z=-i$. Based on how we defined $\Gamma$, $z=i$ is not contained within the closed contour, so we need to find the residue at $z=-i$. We see that\begin{aligned}\text{res}_{-i}f(z) &= \lim_{z\to-i}\frac{\,d}{\,dz}\left[\frac{e^{-2\pi i z\xi}}{(z-i)^2}\right]\\ &=\lim_{z\to-i}\frac{-2\pi i \xi(z-i)^2e^{-2\pi i z\xi}-2(z-i)e^{-2\pi iz\xi}}{(z-i)^4}\\ &=-i\frac{(1+2\pi\xi)e^{-2\pi\xi}}{4}\end{aligned}
Therefore, by the residue theorem, we see that
$\int_{\Gamma}\frac{e^{-2\pi iz\xi}}{(1+z^2)^2} = \int_{-R}^{R}\frac{e^{-2\pi i x\xi}}{(1+x^2)^2}+\int_{C_R}\frac{e^{-2\pi iz\xi}}{(1+z^2)^2} = \frac{\pi}{2}(1+2\pi\xi)e^{-2\pi\xi}$
Observe that $e^{-2\pi iz\xi}\leq 1$ for $\text{Im}z\leq 0$ and $\xi\geq 0$. Thus, as $R\rightarrow\infty$,
$\left|\int_{C_R}\frac{e^{-2\pi iz\xi}}{(1+z^2)^2}\,dz\right|\leq \frac{\pi R}{(R^2+1)^2}$
which goes to zero. Therefore, as $R\rightarrow\infty$, we have
$\int_{-\infty}^{\infty}\frac{e^{-2\pi ix\xi}}{(1+x^2)^2}\,dx=\frac{\pi}{2}(1+2\pi|\xi|)e^{-2\pi|xi|}$
and the proof is complete.

(PS: I just finished posting a solution to POTW 42, so feel free to check that out too!)

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